The potential of the cell $Pt,\underset{\left(1atm\right)}{H_2}|0.1MBCl+0.2MBOH|\left|C{u}^{2+}\left(1M\right)\right|\underset{ }{Cu}$ is 0.88 v at ${25}^{o}C$. BOH is a weak base. $E_{C{u}^{2+}/Cu}^o$ is 0.34 V. If 20 ml 0.1 M BOH is titrated with 0.1 M HCl, the pH at equivalence point is ------    (Given, $\frac{2.303RT}{F}=0.06;\log 2=0.3and\log 5=0.7, K_w={10}^{-14}$ )

Calculation of pH at Equivalence Point

Step 1: At Equivalence Point

20 ml of 0.1 M BOH is titrated with 0.1 M HCl.
At equivalence, all BOH is converted to BOH⁺ (conjugate acid).
Total volume = 20 mL + 20 mL = 40 mL 
[BOH⁺] = (0.1 × 20) / 40 = 0.05 M

Step 2: Calculation from Cell Data

Using the Nernst Equation for the cell:
Pt, H₂ (1 atm) | 0.1 M BCl + 0.2 M BOH || Cu²⁺ (1 M) | Cu
Cell potential: E_cell = 0.88 V
Standard potential for Cu²⁺/Cu: 0.34 V

Nernst: E_cell = E⁰_cell - 0.06 log(Q)

Q = [Cu²⁺][H⁺]² / [BCl][BOH]²

• 0.88 - 0.34 = -0.06 log(Q)
• 0.54 = -0.06 log(Q)
• log(Q) = -9 ⇒ Q = 10^-9
• [H⁺]² / 0.004 = 10^-9
• [H⁺]² = 4 × 10^-12
• [H⁺] = 2 × 10^-6
• pH = -log₁₀(2 × 10^-6) = 5.7

Step 3: pH at Equivalence from Hydrolysis

At equivalence, [BOH⁺] = 0.05 M.
BOH⁺ acts as a weak acid; hydrolysis gives the same effect as above.
Final result reconfirms:

Final Answer: The pH at the equivalence point is 5.7.